Question

A 10 Kg block moving east at 15 m/s collides with a 5 Kg block at rest. The two blocks stick

together and starts moving in an unknown speed.
a. Using the law of conservation of Linear Momentum, calculate the velocity of the
body after collision (V’).
b. Calculate the value of the Kinetic Energy before collision.
c. Calculate the value of the kinetic energy after collision.
d. Is the collision Elastic or Perfectly Inelastic? How did you know?

Answer 1

Step-by-step explanation:

a. According to the law of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision.

Initial momentum = 10 kg * 15 m/s = 150 kg m/s (to the right)

Final momentum = (10 kg + 5 kg) * V'

Therefore, 150 kg m/s = 15 V'

Solving for V', we get:

V' = 10 m/s (to the right)

b. The kinetic energy before the collision can be calculated as:

KE = 1/2 * m1 * v1^2 + 1/2 * m2 * v2^2

where m1 and v1 are the mass and velocity of the 10 kg block, and m2 and v2 are the mass and velocity of the 5 kg block before the collision. Since the 5 kg block is at rest, its kinetic energy is zero.

KE = 1/2 * 10 kg * (15 m/s)^2 = 1125 J

c. After the collision, the two blocks move together with a velocity of 10 m/s to the right. The total kinetic energy after the collision can be calculated as:

KE' = 1/2 * (10 kg + 5 kg) * (10 m/s)^2 = 750 J

d. The collision is perfectly inelastic because the two blocks stick together and move with a common final velocity. This can be confirmed by checking if the kinetic energy is conserved during the collision. Since KE' < KE, we know that some of the initial kinetic energy was lost during the collision, indicating that the collision was not elastic.

Answer 2

A 10 kg block moving east at 15 m/s collides with a 5 kg block at rest. After the collision, the two blocks stick together and move as one. The final velocity of the combined system after the collision is 10 m/s to the east. The change in kinetic energy as a result of the collision is from 1125 J to 750 J. The collision is perfectly inelastic because the two blocks stick together and move as one.

Step-by-step explanation:

a. To find the final velocity of the combined system after the collision, we can use the law of conservation of linear momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision. The equation for momentum is given by: m1v1 + m2v2 = (m1 + m2)v', where m1 and m2 are the masses of the two blocks, v1 and v2 are their respective velocities before the collision, and v' is the velocity of the combined system after the collision. Plugging in the values, we get: (10 kg)(15 m/s) + (5 kg)(0 m/s) = (10 kg + 5 kg)(v'). Solving for v', we find that the velocity of the combined system after the collision is 10 m/s to the east.

b. The kinetic energy before the collision can be calculated using the equation: KE1 = 0.5m1v12 + 0.5m2v22, where KE1 is the initial kinetic energy, m1 and m2 are the masses of the two blocks, and v1 and v2 are their respective velocities before the collision. Plugging in the values, we get: KE1 = 0.5(10 kg)(15 m/s)2 + 0.5(5 kg)(0 m/s)2 = 1125 J.

c. The kinetic energy after the collision can be calculated using the equation: KE2 = 0.5(m1 + m2)v'2, where KE2 is the final kinetic energy, m1 and m2 are the masses of the two blocks, and v' is the velocity of the combined system after the collision. Plugging in the values, we get: KE2 = 0.5(10 kg + 5 kg)(10 m/s)2 = 750 J.

d. The collision is perfectly inelastic because the two blocks stick together after the collision and move as one. This can be deduced from the fact that the final velocity of the combined system is less than the initial velocity of block A. In an elastic collision, the objects would bounce off each other and separate.